Problem: Rewrite the function by completing the square. $f(x)=x^{2}-4x-26$ $f(x)=(x+$
Solution: We want to complete $x^2{-4}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-4}}{2}\right)^2={4}$ to it: $x^2{-4}x+{4}=(x-2)^2$ In order to keep the expression equivalent, we add and subtract ${4}$, not forgetting the expression's constant term, $-26$ : $\begin{aligned} f(x)&=x^2-4x-26 \\\\ &=x^2-4x+{4}-26-{4} \\\\ &=(x-2)^2-26-4 \\\\ &=(x-2)^2-30 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 2)^2 -30$ This is equivalent to $f(x)=(x+{-2})^2-30$